Java 启动活动并上载数据库时出错
当我点击一个按钮时,我开始一个新的活动,名为Profilo.java。我的代码是: 包com.example.prenotazione_esame 导入android.app.Activity;导入android.app.AlertDialog;进口 android.content.DialogInterface;导入android.database.Cursor; 导入android.database.sqlite.SQLiteDatabase;导入android.os.*; 导入android.view.Menu;导入android.view.MenuItem;进口 android.widget.Button;导入android.widget.EditText;进口 android.view.view;导入android.view.view.OnClickListener;进口 android.widget.ToastJava 启动活动并上载数据库时出错,java,android,database,android-activity,android-sqlite,Java,Android,Database,Android Activity,Android Sqlite,当我点击一个按钮时,我开始一个新的活动,名为Profilo.java。我的代码是: 包com.example.prenotazione_esame 导入android.app.Activity;导入android.app.AlertDialog;进口 android.content.DialogInterface;导入android.database.Cursor; 导入android.database.sqlite.SQLiteDatabase;导入android.os.*; 导入androi
public class Profilo extends Activity {
private LoginDataBase dbLogin;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
int id = getIntent().getExtras().getInt("id");
setContentView(R.layout.profilo);
SQLiteDatabase db = dbLogin.getWritableDatabase();
}
}
这是LoginDataBase.java
public class LoginDataBase extends SQLiteOpenHelper {
private static final String DB_NAME="Login_DB";
private static final int DB_VERSION=1;
public LoginDataBase(Context context){
super(context,DB_NAME,null,DB_VERSION);
}
@Override
public void onCreate(SQLiteDatabase db) {
String sql="";
sql+= "CREATE TABLE T_LOGIN (";
sql+= " _id_LOGIN INTEGER PRIMARY KEY AUTOINCREMENT,";
sql+= " USERNAME TEXT NOT NULL,";
sql+= " PASSWORD TEXT NOT NULL";
sql+=")";
db.execSQL(sql);
String sql2="";
sql2+= "CREATE TABLE T_PROFILO (";
sql2+= " _id_PROFILO INTEGER PRIMARY KEY AUTOINCREMENT,";
sql2+= " NOME TEXT NOT NULL,";
sql2+= " COGNOME TEXT NOT NULL,";
sql2+= " ETA TEXT NOT NULL,";
sql2+= " SESSO TEXT(1) NOT NULL,";
sql2+= " CODICE_FISCALE TEXT NOT NULL,";
sql2+= " CITTA TEXT NOT NULL,";
sql2+= " INDIRIZZO TEXT NOT NULL,";
sql2+= " TELEFONO TEXT NOT NULL,";
sql2+= " _id_L INTEGER NOT NULL,";
sql2+= " FOREIGN KEY(_id_L) REFERENCES T_LOGIN(_id_LOGIN)";
sql2+= ")";
db.execSQL(sql2);
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
// Aggiornamento delle tabelle
}
}
问题是当我尝试获取数据库数据库数据库时。我的应用程序崩溃了。
问题出在哪里?您需要对dbLogin进行初始化
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
int id = getIntent().getExtras().getInt("id");
setContentView(R.layout.profilo);
///NEW
LoginDataBase dbLogin = new LoginDataBase (this, "NAMEOFDATABASE", null, 1);
///END OF NEW
SQLiteDatabase db = dbLogin.getWritableDatabase();
}
}
您的类LoginDataBase必须具有与此类似的构造函数:
public class LoginDataBase extends SQLiteOpenHelper {
private static final String DB_NAME="Login_DB";
private static final int DB_VERSION=1;
public LoginDataBase (Context contexto, String name,
CursorFactory factory, int version) {
super(contexto, name, factory, version);
}
或者您也可以更改此设置并继续使用构造函数:
///NEW
LoginDataBase dbLogin = new LoginDataBase (this);
///END OF NEW
这次我有一个错误:构造函数登录数据库未定义。什么意思?谢谢。这意味着类LoginDataBase有一个不同于“=new LoginDataBase(这是“NAMEOFDATABASE”,null,1)”的构造函数,或者没有任何构造函数。使两者具有相同的字段。P.D.Post LoginDataBase Constructor或Post LoginDataBase.java。非常感谢。